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How can one calculate the mass of air inside a tyre, given a particular tyre size; a pressure, in $kPa = \frac{1000kg}{m\cdot s^2}$; and assuming room temperature, and normal air composition? I can't quite work out what part of the equations I'm missing to remove the $s^2$ component.

I realise that surface/volume ratio is important, but for the purposes and approximation to a torus would be fine. For example, assuming a 700cx25 bicycle tyre, we might assume a torus where the diameter between centre of the two cross-section circles is about 630mm, and the diameter of the circles themselves is about 30mm. Let's assume a tyre pressure of 500kPa (~73psi).

Rough volume would be $630$mm$\times\pi \times \pi (15$mm$)^2 = 1.4\times10^6$mm$^3 = 0.0014$m$^3$

Rough surface area: $630$mm$\times\pi \times 30$mm$\times\pi = 1.87\times10^5$mm$^2 = 0.187$m$^2$

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up vote 3 down vote accepted

You need an equation for the density of the gas as a function of temperature and pressure. Assuming the tyre is full of air, this is reasonably close to an ideal gas so the molar volume is given by:

$$V_m = \frac{RT}{P}$$

where R is the ideal gas constant and and the average molecular weight of air (20% O$_2$, 80% N$_2$) is about 28.8. From this you can work out the density at the pressure and temperature of your car tyre, and since you know the volume this immediately gives you the mass.

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Or you could just use the experimental density of air at room TP is around 1.2 kg/m^3. Together with how many Bar you inflate your tire to – Martin Beckett Aug 1 '12 at 12:58

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